codeforces 1140D(区间dp/思惟题)

D. Minimum Triangulation

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given a regular polygon with nn vertices labeled from 11 to nn in counter-clockwise order. The triangulation of a given polygon is a set of triangles such that each vertex of each triangle is a vertex of the initial polygon, there is no pair of triangles such that their intersection has non-zero area, and the total area of all triangles is equal to the area of the given polygon. The weight of a triangulation is the sum of weigths of triangles it consists of, where the weight of a triagle is denoted as the product of labels of its vertices.

Calculate the minimum weight among all triangulations of the polygon.

Input

The first line contains single integer nn (3≤n≤5003≤n≤500) — the number of vertices in the regular polygon.

Output

Print one integer — the minimum weight among all triangulations of the given polygon.

Examples

input

Copy

3

output

Copy

6

input

Copy

4

output

Copy

18

Note

According to Wiki: polygon triangulation is the decomposition of a polygonal area (simple polygon) PP into a set of triangles, i. e., finding a set of triangles with pairwise non-intersecting interiors whose union is PP.

In the first example the polygon is a triangle, so we don't need to cut it further, so the answer is 1⋅2⋅3=61⋅2⋅3=6.

In the second example the polygon is a rectangle, so it should be divided into two triangles. It's optimal to cut it using diagonal 1−31−3 so answer is 1⋅2⋅3+1⋅3⋅4=6+12=181⋅2⋅3+1⋅3⋅4=6+12=18.

 题意：给你一个有n个顶点的正多边形，顶点编号从1-n，如今你要把这个正多边形划分为三角形，并且每一个三角形的面积都不相交，三角形的花费定义为三角形顶点编号的乘积，问花费的最少代价。

 

思路：简单的区间dp，对于dp[i][j]的求解，咱们只要以i，j为底边，枚举顶点k（i<k<j），划分出三角形（i，j，k），而后咱们就将要求的值分为dp[i][k]+dp[k][j]+三角形（i，j，k）的花费，找到花费最小的值就能够了。

若是敏感的话应该能够发现对于正多边形的划分，就是划分为（1，2 ，3），（1，3，4），（1，4，5）。。。。，（1，n-1，n）时它的花费是最小的，那么最后的答案就是2*3+3*4+4*5+.......+（n-1）*n。

 

#include<cstdio> #include<algorithm>
using namespace std; const int INF=1e9; int dp[510][510]; int main(){ int n; scanf("%d",&n); for(int j=2;j<=n-1;j++){ for(int i=1;i+j<=n;i++){ dp[i][i+j]=INF;//找最小值，先初始化为无穷大 
            for(int k=i+1;k<i+j;k++){ dp[i][i+j]=min(dp[i][i+j],dp[i][k]+dp[k][i+j]+i*(i+j)*k); } } } printf("%d\n",dp[1][n]); return 0; } 

View Code

#include<cstdio> #include<algorithm>
using namespace std; const int INF=1e9; int main(){ int n; scanf("%d",&n); int ans=0; for(int i=3;i<=n;i++) ans+=i*(i-1); printf("%d\n",ans); return 0; }

View Code